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Mariana [72]
4 years ago
15

If it takes 43.2 mL of 0.1 M NaOH to neutralize a 50 mL HCl solution, how many moles of NaOH were added to the HCl solution?

Chemistry
1 answer:
OleMash [197]4 years ago
8 0

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Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid wate
allochka39001 [22]

Answer:

a) 2NaOH+H2SO4→Na2SO4+2H2O2NaOH+H2SO4→Na2SO4+2H2O

b) Số phân tử NaOH : Số phân tử H2SO4 = 2:1

Số phân tử NaOH : Số phân tử Na2SO4 = 2:1

Số phân tử NaOH : Số phân tử H2O = 2:2

Explanation:

8 0
2 years ago
I only neee help on 1 through 6 please help me the questions are in the picture?!??!!!!!?!
Galina-37 [17]

Answer:

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4 0
3 years ago
How many moles of O are present in 2.9 mol of Al2(SO4)3 ?
ad-work [718]

Answer:

34.8 moles of O

Explanation:

First we break down Al2(SO4)3 and look at SO4 only. In this compound, there are already 4 moles of O. Then you see that there are 3 moles of (SO4), meaning there are 3*4 moles of O = 12 moles in the entire compound. Then you look at the question, where it's asking for 2.9 moles of the entire Al2(SO4)3 compound, so you multiply 2.9*12 = 34.8 total moles of O.

6 0
4 years ago
150.0 grams of an isotope with a half-life of 36.0 hours is present at time zero. How long will it take
leonid [27]

Given :

Initial mass , a = 150 gm .

Half time ,

t_{\dfrac{1}{2}}=36 \ hours.

Final mass , x = 18.75 gm .

To Find :

The time taken to decay 18.75 gm .

Solution :

We know , time taken is given by :

t=\dfrac{1}{k}\times ln(\dfrac{a}{a-x})

Here , k is a constant given by :

k=\dfrac{0.693}{t_{\dfrac{1}{2}}}\\\\\\k=\dfrac{0.693}{36}

Putting all given value in above equation :

We get :

t=\dfrac{36}{0.693}\times ln(\dfrac{150}{150-18.75})\\\\t=6.94\ hours

Therefore , time taken is 6.94 hours .

Hence , this is the required solution .

5 0
4 years ago
Which expression gives the standard enthalpy<br> change of formation for methanol?
ser-zykov [4K]

Answer:

25

Explanation:

bec

6 0
3 years ago
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